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Momentum problem with collision

  1. Feb 25, 2007 #1

    lzh

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    1. The problem statement, all variables and given/known data
    Assume an elastic collision (ignoring friction
    and rotational motion).
    A queue ball initially moving at 3.7 m/s
    strikes a stationary eight ball of the same size
    and mass. After the collision, the queue ball's
    final speed is 2.1 m/s at an angle of theta with
    respect to its original line of motion.

    [​IMG]
    Find the eight ball's speed after the colli-
    sion. Answer in units of m=s.
    2. Relevant equations
    p=mv
    possibly energy equations


    3. The attempt at a solution
    This problem would've been alot easier if the problem provided theta's value. Without theta i only managed to get to this:
    3.7=2.1cos(theta)+Vx
    where Vx is the x component of the 8 ball after collision
    0=2.1sin(theta)+Vy
    where Vy is the y component of the 8 ball after collision
    should i find theta first? or can I find one of the component's value first and then find theta?
     
  2. jcsd
  3. Feb 25, 2007 #2

    radou

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    Homework Helper

    First of all, you missed a sign.
     
  4. Feb 25, 2007 #3

    lzh

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    that doesnt matter does it?
    just because Vy is negative you dont need to do this:
    0=2.1sin(theta)-Vy
    since that would end up having Vy's value as positive
     
  5. Feb 25, 2007 #4
    Vy is positive, and that is why you have to put the negative sign in front.

    Although you haven't explicitly written it down, it's the components of the velocity vectors that you are equating.

    So, you're second equation is something like
    0j=2.1sin(theta)j-Vyj
     
  6. Feb 25, 2007 #5

    lzh

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    j is the mass of the ball?
    if so I left them out because they cancel
     
  7. Feb 25, 2007 #6
    By j, I meant a unit-vector in the direction parallel to the y-axis.

    0j=2.1sin(theta)j-Vyj
    => 0j=(2.1sin(theta)-Vy)j

    Since the vectors are equal, so are their magnitudes, and that's how you get
    0=2.1sin(theta)-Vy
     
  8. Feb 25, 2007 #7

    lzh

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    oh, but I dont really see how that helps me in the problem
     
  9. Feb 25, 2007 #8
    hey guys may b u shud consider forming equs fom the line of collision.....tats how they did in our class....im not v comfertable wit it either, but ill get ma note book if u want
     
  10. Feb 25, 2007 #9

    lzh

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    you mean breaking it into components? and having equation for x direction and y direction?
     
  11. Feb 25, 2007 #10
    well, in a similar prob, v got the first equs by e=v(seperation)/v(approach)...the other relations were used on the basis of LOC
     
  12. Feb 25, 2007 #11
    yea ...sumthing like tat...shud i get ma notebook?
     
  13. Feb 25, 2007 #12

    lzh

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    this problem supposedly should use momentum formulae. And i dont think I've ever seen your way of doing it? I sorta dont get it... what exactly does e stand for here and what is LOC.
    yeah plz get your notebook
     
  14. Feb 25, 2007 #13
    e=co-eff of restitution, n LOC is line of collision, , it says here newtons law of collision can only b applied along the loc
     
  15. Feb 25, 2007 #14
    ummm...u shud know tat im living in bombay, so even if the laws of physics dont change, the methods of attacking a ques do change ....n im currently trainig for iit entrance exam...so, idk wht level u r at
     
  16. Feb 25, 2007 #15

    lzh

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    ah now i sort of get it, though they never taught about coefficient of restitution in school yet. So basically once i find the coeff of restitution all i need to do is subtract it by 2.1 m/s.
     
  17. Feb 25, 2007 #16
    now, for any spherrical bodies, loc is the line joining the centres of the two coliding bodies......u take the components of the resulting velocities along the loc, get the equs n solve......now i can giv u the solution inna while, but my personal advise wud b ,dont listen to me atall if the concept of loc is entirely new to u.....but frankly, i dont c how the prob can b tackled otherwise
     
  18. Feb 25, 2007 #17
    not quite, u on msn?
     
  19. Feb 25, 2007 #18

    lzh

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    but there are 2 unknowns for this case, the velocity of 8 ball after collision and the coeff of restitution
     
  20. Feb 25, 2007 #19
    dear, u just said tat its a elastic collision, n to avoid friction n all, u ment a perfectly elastic collision, in this case, e=1, always
     
  21. Feb 25, 2007 #20
    whcih brings us to the eqn u already hav, u1= v1cos(theta)- v2....the udder eqs, i recommend u get aquainted wit the concept of loc
     
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